What is the time complexity of Find Indices With Index and Value Difference II?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Find Indices With Index and Value Difference II?

This problem uses Array, Two Pointers. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find Indices With Index and Value Difference II?

Insight 1: The problem can be solved in linear time by maintaining a running track of the minimum and maximum values encountered within a sliding window defined by `indexDifference`. Insight 2: Iterate through the array starting from index `indexDifference`. For each current index `j`, check if a suitable index `i` exists within the range `[0, j - indexDifference]` using the stored minimum and maximum values seen so far.

Which companies ask LeetCode 2905. Find Indices With Index and Value Difference II in interviews?

This problem is frequently asked at Paytm.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find Indices With Index and Value Difference II - LeetCode 2905 Solution

Find Indices With Index and Value Difference II - Complete Solution Guide

Find Indices With Index and Value Difference II is LeetCode problem 2905, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed integer array nums having length n , an integer indexDifference , and an integer valueDifference . Your task is to find two indices i and j , both in the range [0, n - 1] , that satisfy the following conditions: abs(i - j) >= indexDifference , and abs(nums[i] - nums[j]) >= valueDifference Return an integer array answer , where answer = [i, j] if there are two such indices , and answer = [-1, -1] otherwise . If there are multiple choices for the two indices, return any o

Detailed Explanation

The problem requires us to find two indices, `i` and `j`, within an array `nums` that satisfy two conditions: the absolute difference between the indices (`|i - j|`) must be greater than or equal to `indexDifference`, and the absolute difference between the corresponding values in the array (`|nums[i] - nums[j]|`) must be greater than or equal to `valueDifference`. The goal is to return an array `[i, j]` representing such indices if they exist. If no such pair of indices exists, return `[-1, -1]`. The indices `i` and `j` can be the same.

Solution Approach

The provided solution uses a single pass through the array `nums` to find the indices `i` and `j` that satisfy the given conditions. It maintains `min_idx` and `max_idx` to keep track of the indices of the minimum and maximum values seen so far within the sliding window of size `indexDifference`. For each index `j` from `indexDifference` to the end of the array, it checks if the value at `j` satisfies the `valueDifference` condition with either the minimum or maximum value found so far in the window defined by `j - indexDifference`. If a suitable pair is found, it returns the indices immediately. Otherwise, it updates `min_idx` and `max_idx` to reflect any new minimum or maximum values encountered in the previous window element.

Step-by-Step Algorithm

Step 1: Initialize `min_idx` and `max_idx` to 0. These will store the indices of the minimum and maximum values encountered so far.
Step 2: Iterate through the array `nums` starting from `j = indexDifference` to the end of the array.
Step 3: In each iteration, calculate `i = j - indexDifference` to represent the index at the beginning of the current window.
Step 4: Update `min_idx` and `max_idx` if `nums[i]` is smaller than `nums[min_idx]` or larger than `nums[max_idx]`, respectively.
Step 5: Check if `nums[max_idx] - nums[j] >= valueDifference`. If it is, return `[max_idx, j]`.
Step 6: Check if `nums[j] - nums[min_idx] >= valueDifference`. If it is, return `[min_idx, j]`.
Step 7: If the loop completes without finding a suitable pair, return `[-1, -1]`.

Key Insights

Insight 1: The problem can be solved in linear time by maintaining a running track of the minimum and maximum values encountered within a sliding window defined by `indexDifference`.
Insight 2: Iterate through the array starting from index `indexDifference`. For each current index `j`, check if a suitable index `i` exists within the range `[0, j - indexDifference]` using the stored minimum and maximum values seen so far.
Insight 3: The problem implicitly asks for *any* valid solution, allowing us to terminate the search as soon as one is found. This avoids the need to find all pairs and then select one according to some other criterion (which doesn't exist in the problem statement).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: Paytm.

Find Indices With Index and Value Difference II | LeetCode 2905

Medium

Array Two Pointers

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
        int min_idx = 0;
        int max_idx = 0;
        
        for (int j = indexDifference; j < nums.length; j++) {
            int i = j - indexDifference;
            
            if (nums[i] < nums[min_idx]) {
                min_idx = i;
            }
            if (nums[i] > nums[max_idx]) {
                max_idx = i;
            }
            
            if (nums[max_idx] - nums[j] >= valueDifference) {
                return new int[]{max_idx, j};
            }
            
            if (nums[j] - nums[min_idx] >= valueDifference) {
                return new int[]{min_idx, j};
            }
        }
        
        return new int[]{-1, -1};
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2905: Find Indices With Index and Value Difference II - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

Paytm

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2905

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `min_idx` and `max_idx` to 0. These will store the indices of the minimum and maximum values encountered so far.
2
Step 2: Iterate through the array `nums` starting from `j = indexDifference` to the end of the array.
3
Step 3: In each iteration, calculate `i = j - indexDifference` to represent the index at the beginning of the current window.
4
Step 4: Update `min_idx` and `max_idx` if `nums[i]` is smaller than `nums[min_idx]` or larger than `nums[max_idx]`, respectively.
5
Step 5: Check if `nums[max_idx] - nums[j] >= valueDifference`. If it is, return `[max_idx, j]`.
6
Step 6: Check if `nums[j] - nums[min_idx] >= valueDifference`. If it is, return `[min_idx, j]`.
7
Step 7: If the loop completes without finding a suitable pair, return `[-1, -1]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the main loop iterates through the array once. The space complexity is constant because the algorithm only uses a fixed number of variables, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs of indices (i, j) and check if they satisfy the conditions.

Time: O(n^2), where n is the length of the array `nums`.Space: O(1)

Trade-offs: The brute force approach is simple to understand but inefficient for large arrays. The provided solution is much more efficient with its O(n) time complexity. The brute force approach should only be considered if the input size is extremely small.

✨

Key Insights

1Insight 1: The problem can be solved in linear time by maintaining a running track of the minimum and maximum values encountered within a sliding window defined by `indexDifference`.
2Insight 2: Iterate through the array starting from index `indexDifference`. For each current index `j`, check if a suitable index `i` exists within the range `[0, j - indexDifference]` using the stored minimum and maximum values seen so far.
3Insight 3: The problem implicitly asks for *any* valid solution, allowing us to terminate the search as soon as one is found. This avoids the need to find all pairs and then select one according to some other criterion (which doesn't exist in the problem statement).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: `indexDifference = 0` and `valueDifference = 0`. The solution correctly handles this case because the loop starts at `j = 0` and the difference check allows for equality, so any index paired with itself will be valid.
Edge case 2: No pair of indices satisfies the conditions. The solution correctly returns `[-1, -1]` after iterating through the entire array without finding a suitable pair.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `indexDifference` is 0, leading to incorrect results.

✗Common mistake 2: Incorrectly updating `min_idx` and `max_idx` or not updating them at all. This leads to using outdated minimum and maximum values, potentially missing valid solutions.

✗Common mistake 3: Returning the indices in the wrong order. While the problem does not specify an order constraint, ensuring consistent behavior is good practice.

✗Common mistake 4: Not considering the possibility that `i` and `j` could be the same when `indexDifference` and `valueDifference` are zero.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing stock prices to find instances where a significant price difference occurs after a certain time delay. `nums` represents the stock prices, `indexDifference` is the time delay, and `valueDifference` is the minimum price change required.

🏦Real-world scenario 2: Detecting anomalies in sensor data. `nums` represents the sensor readings, `indexDifference` is the minimum time between anomalies, and `valueDifference` is the minimum difference in sensor readings that constitutes an anomaly.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Indices With Index and Value Difference II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?